To solve the equation [tex]\(2x^2 = -10x + 12\)[/tex], we will first rearrange it to standard quadratic form, which is [tex]\(ax^2 + bx + c = 0\)[/tex].
Given equation:
[tex]\[
2x^2 = -10x + 12
\][/tex]
Rearrange to:
[tex]\[
2x^2 + 10x - 12 = 0
\][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = -12\)[/tex].
The solutions to this equation can be found using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute [tex]\(a = 2\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = -12\)[/tex] into the formula:
[tex]\[
x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 2 \cdot (-12)}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{-10 \pm \sqrt{100 + 96}}{4}
\][/tex]
[tex]\[
x = \frac{-10 \pm \sqrt{196}}{4}
\][/tex]
[tex]\[
x = \frac{-10 \pm 14}{4}
\][/tex]
Now, solve for the two possible values of [tex]\(x\)[/tex]:
1. When [tex]\( x = \frac{-10 + 14}{4} \)[/tex]:
[tex]\[
x = \frac{4}{4}
\][/tex]
[tex]\[
x = 1
\][/tex]
2. When [tex]\( x = \frac{-10 - 14}{4} \)[/tex]:
[tex]\[
x = \frac{-24}{4}
\][/tex]
[tex]\[
x = -6
\][/tex]
Thus, the solutions to the equation [tex]\(2x^2 = -10x + 12\)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = -6
\][/tex]
From the given options, the correct answers are:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = -6 \)[/tex]
